Question

Find the area of the region\({(x,y):y^2≤4x,4x^2+4y^2≤9}\)

Answer 1

The correct answer is:\(∫^{\frac{1}{2}}_02\sqrt{x}dx+∫^{\frac{3}{2}}_{\frac{1}{2}}\frac{1}{2}\sqrt{(3)^2-(2x)^2}dx\)
The area bounded by the curves,\({(x,y):y^2≤4x,4x^2+4y^2≤9}\),is represented as

The points of intersection of both the curves are\((\frac{1}{2},\sqrt{2})\)and\((\frac{1}{2},-\sqrt{2}).\)
The required area is given by OABCO.
It can be observed that area OABCO is symmetrical about \(x\)-axis.
\(∴ Area\,\, OABCO=2\times Area\,\, OBC\)
\(Area\,\, OBCO=Area\,\, OMC+Area\,\, MBC\)
\(=∫^{\frac{1}{2}}_02\sqrt{x}dx+∫^{\frac{3}{2}}_{\frac{1}{2}}\frac{1}{2}\sqrt{9-4x^2}dx\)
\(=∫^{\frac{1}{2}}_02\sqrt{x}dx+∫^{\frac{3}{2}}_{\frac{1}{2}}\frac{1}{2}\sqrt{(3)^2-(2x)^2}dx\)